Bulk Modulus & amp; Work per Unit Volume

Edwin Ogie Library

Bulk Modulus & Work per Unit Volume

Detailed Note on Bulk Modulus

Bulk Modulus (K) measures a material's resistance to uniform compression. It is defined as the ratio of the change in pressure to the resulting fractional change in volume.

K = ΔP / (ΔV / V)

K = - (ΔP · V) / ΔV

The negative sign indicates that an increase in pressure (positive ΔP) produces a decrease in volume (negative ΔV), making K positive. Units: Pascal (Pa = N/m²).

Work done (Elastic potential energy) per unit volume

For small elastic deformations (linear region), the energy stored per unit volume (elastic energy density) is equal to the area under the stress–strain curve:

U = ½ × (stress) × (strain) = ½ × (F/A) × (ΔL/L)

Applications & notes

• Speed of sound in a medium: v = √(K/ρ).
• Liquids typically have large bulk moduli (nearly incompressible) compared to gases.
• K is positive in standard sign convention and reported as a magnitude for material properties.

10 JAMB Worked Examples

Example 1

Question: A cube has volume V = 1.00×10⁻³ m³. When pressure is increased by ΔP = 2.00×10⁵ Pa its volume decreases by ΔV = -1.00×10⁻⁶ m³. Find K.

Solution:

ΔV/V = (-1.00×10⁻⁶)/(1.00×10⁻³) = -1.00×10⁻³

K = ΔP / (ΔV/V) = 2.00×10⁵ / 1.00×10⁻³ = 2.00×10⁸ Pa

Example 2

Question: A body of volume V = 0.020 m³ decreases by 2.0×10⁻⁵ m³ under ΔP = 1.0×10⁷ Pa. Find K.

Solution:

ΔV/V = (2.0×10⁻⁵)/0.020 = 1.0×10⁻³

K = 1.0×10⁷ / 1.0×10⁻³ = 1.0×10¹⁰ Pa

Example 3

Question: A wire (L = 2.00 m, A = 1.00×10⁻⁶ m²) is stretched by F = 5.00 N producing ΔL = 2.00×10⁻³ m. Find energy per unit volume U.

Solution:

Stress σ = F/A = 5.00 / 1.00×10⁻⁶ = 5.00×10⁶ Pa

Strain ε = ΔL / L = 2.00×10⁻³ / 2.00 = 1.00×10⁻³

U = ½ σ ε = 0.5 × 5.00×10⁶ × 1.00×10⁻³ = 2.50×10³ J/m³

Example 4

Question: Given K = 2.20×10⁹ Pa and ΔP = 1.10×10⁶ Pa. Find ΔV/V.

Solution:

ΔV/V = ΔP / K = 1.10×10⁶ / 2.20×10⁹ = 5.00×10⁻⁴

Example 5

Question: A sphere of volume 0.500 m³ reduces by 5.00×10⁻⁵ m³ under ΔP. If K = 1.00×10⁹ Pa, find ΔP (magnitude).

Solution:

ΔV/V = (-5.00×10⁻⁵)/0.500 = -1.00×10⁻⁴

ΔP = K (ΔV/V) = 1.00×10⁹ × (-1.00×10⁻⁴) ⇒ magnitude = 1.00×10⁵ Pa

Example 6

Question: Steel: K = 1.60×10¹¹ Pa, ρ = 7800 kg/m³. Find speed of sound v.

Solution:

v = √(K/ρ) = √(1.60×10¹¹ / 7800) ≈ 4.529×10³ m/s

Example 7

Question: If U = 1000 J/m³ and stress σ = 2.00×10⁶ Pa, find strain ε.

Solution:

ε = 2U / σ = 2×1000 / 2.00×10⁶ = 1.00×10⁻³

Example 8

Question: Liquid: K = 2.20×10⁹ Pa, ρ = 1000 kg/m³. Compute v.

Solution:

v = √(K/ρ) = √(2.20×10⁹ / 1000) ≈ 1.483×10³ m/s

Example 9

Question: Block V = 0.100 m³ reduces to 0.0999 m³ under ΔP = 5.00×10⁶ Pa. Find K.

Solution:

ΔV = -1.00×10⁻⁴ m³ ⇒ ΔV/V = -1.00×10⁻³

K = 5.00×10⁶ / 1.00×10⁻³ = 5.00×10⁹ Pa

Example 10

Question: Spring k = 200 N/m stretched by x = 0.10 m. L = 0.50 m, A = 1.00×10⁻⁴ m². Find U.

Solution:

Es = ½ k x² = 1.00 J; V ≈ A L = 5.00×10⁻⁵ m³; U = Es / V = 2.00×10⁴ J/m³

JAMB CBT Quiz on Bulk Modulus & Energy Density

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